For x ∈ ( 0 , 2 5 π ) , define f ( x ) = ∫ 0 x t sin t d t . Then f has

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Mathematics - Definite Integration Question with Solution | TestHub

MathematicsDefinite IntegrationDerivatives (Newton- Leibnitz)Hard2 minPYQ_2011

MathematicsHardsingle choice

For $x \in (0, \frac{5 π}{2})$ , define $f (x) = \int_{0}^{x} t sin t d t$ . Then $f$ has

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$f^{'} (x) = x sin x$ Given $x \in (0, \frac{5 π}{2})$ $f^{'} (x)$ changes sign from +ve to -ve at $π$ $f^{'} (x)$ changes sign from -ve to +ve at $2 π$ $f$ has local max at $π$ , local min at $2 π$

Stream:JEESubject:MathematicsTopic:Definite IntegrationSubtopic:Derivatives (Newton- Leibnitz)

⏱ 2mℹ️ Source: PYQ_2011

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